Question

The following program is executed by 2 processes p and q concurrently:

x = x + 1
The variable x is shared by the two processes.
The instruction x = x + 1 is compiled into 3 machine instructions to load x into a register Rp, increment the register value, and store the register back into x.
load Rp, x
add Rp, 1
store Rp, x
a. If x is initially 2, determine the expected value of x when both p and q terminate.
b. In the absence of any synchronization primitives, program interleaving can result in a loss of 1 or more of the updates. Show one sequence instruction execution where the final value of x is less than the expected value 4.

Answer 1

In the absence of synchronization, two concurrent processes updating the same variable can lead to a race condition with a final value less than expected. If 'x' starts at 2 and both processes increment 'x', the expected value is 4, but interleaved instructions can result in a final value of 3.

Step-by-step explanation:

The question addresses a scenario in concurrent programming where two processes p and q are updating a shared variable x concurrently without any synchronization mechanisms. This situation can lead to a race condition.

a. If x is initially 2, and both processes p and q terminate after executing the instruction x = x + 1 successfully without any interference, the expected value of x would be 4.

b. However, without synchronization, interleaving of instructions can occur, leading to a race condition and therefore the final value of x could be less than the expected 4. An example sequence that results in a final value of x being 3 is:

• Process p: load Rp, x (Rp is now 2)
• Process q: load Rq, x (Rq is also now 2)
• Process p: add Rp, 1 (Rp is now 3)
• Process q: add Rq, 1 (Rq is now 3)
• Process p: store Rp, x (x is now 3)
• Process q: store Rq, x (x remains 3)

In this sequence of operations, the update by process p is overwritten by the update of process q because they both loaded the same initial value of x.